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3 years ago

Concentration and ion moles are equal only when volume is what ?

Chemistry

1 answer:

siniylev [52]3 years ago

3 0

Concentration and ion moles are equal only when volume is 1. Of 1.0 x 10-6 and 1.0 x 10-4, the larger number is 1.

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What is the outcome of all chemical changes when two substances are combined?

wel

Answer:

Explanation:

Because when you combine chemical substances they create a new substance unlike physical substances .

3 0

3 years ago

Idk how to do this at all!!!

sergey [27]

Hello there!

I am not really sure which part you are talking about but let me answer some of them.

First, you need to read the information they gave you about glucose etc..

13- What is produced in photosynthesis?
"Photosynthesis is essential for all life on earth, because it provides food and oxygen."

Thus, the answer is: Photosynthesis produced food and oxygen.

14- What is the glucose used for?
"The glucose produced is used by the plant for energy and growth."

Thus, the answer is: The glucose used for energy and growth of the plants.

15- What is the oxygen used for?
"The oxygen produced is released into the air for us to breath."

Thus, the answer is: The oxygen used for us to breath.

I hope this helps!

Let me know if you have questions about the answer.

3 0

3 years ago

The complex [Ni(CN)4]2- is diamagnetic and the complex [NiCl4]2- is paramagnetic. What can you conclude about their molecular ge

quester [9]

Answer:

[Ni(CN)4]2- square planar

[NiCl4]2- tetrahedral

Explanation:

For a four coordinate complex such as [Ni(CN)4]2- and [NiCl4]2-, we can decide its geometry by closely considering its magnetic properties. Both of the complexes are d8 complexes which could be found either in the tetrahedral or square planar crystal field depending on the nature of the ligand.

CN^- being a strong field ligand leads to the formation of a square planar diamagnetic d8 complex of Ni^2+. Similarly, Cl^- being a weak field ligand leads to the formation a a tetrahedral paramagnetic d8 complex of Ni^+ hence the answer given above.

7 0

3 years ago

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 3.6-L bulb, then filled

aleksandrvk [35]

Answer:

The gas was N₂

Explanation:

V = 3.6L

P = 2.0 atm

T = 24.0°C = 297K

R = 0.0821 L.atm/K.mol

m = 8.3g

M = molar mass = ?

Using ideal gas equation;

PV = nRT

n = no. Of moles = mass / molar mass

n = m/M

PV = m/M * RT

M = mRT / PV

M = (8.3*0.0821*297) / (2.0*3.6)

M = 28.10

Since X is a diatomic molecule

M = 28.10 / 2 = 14.05 g/mol

M = Nitrogen

X = N₂

5 0

2 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago